Question

The radius of a hydrogen atom in the ground state is $0.53\mathring{A}$. What will be the radius of $L{i}^{2+}$ in the ground state?

• A. $1.06\mathring{A}$
• B. $0.265\mathring{A}$
• C. $0.17\mathring{A}$
• D. $0.53\mathring{A}$

Answer 1

Answer: (C)

$0.17\mathring{A}$

The radii of the $n^{th}$ stationary state for a hydrogen-like specie is expressed as :

$r_n=\frac{n^2{a}_0}{Z}$

where $Z=$atomic number and $a_0=$ radius of Bohr orbit.

For ground state of hydrogen atom, $n=1$ and $Z=1$

thus for Hydrogen $r_1\left(H\right)=a_0=0.53\mathring{A}$ (given)

Thus the radius of the first stationary state of hydrogen atom i.e. in ground state, called the Bohr orbit, is $0.53\mathring{A}$.

for ${}_{3}L{i}^{2+}$, Z=3 and ground state means $n=1$

thus $r_1\left(L{i}^{2+}\right)=\frac{1^2\times 0.53}{3}$

or $r_1\left(L{i}^{2+}\right)=0.176\mathring{A}$